Prove that $\min \{a^7,b^3,c^2,1\} \le abc$

Question

If $a,b,c$ are three non-negative real numbers, prove that:

$$\min \{a^7,b^3,c^2,1\} \le abc$$

I tried several ideas with the geometric mean:

$$\min \{a^7,b^3,c^2,1\} \leq \sqrt[4]{a^7b^3c^2}$$

but $\sqrt[4]{a^7b^3c^2} \leq abc$ is not true. I tried with other means but nothing significant. I think these ideas fail because the inequality is not homognenous.

Answer 1

For the sake of contradiction, assume that:

$$\min\{a^7,b^3,c^2,1\} > abc$$

Then:

$$a^7 > abc \Rightarrow a > (abc)^{\frac{1}{7}}$$

$$b^3 > abc \Rightarrow b > (abc)^{\frac{1}{3}}$$

$$c^2 > abc \Rightarrow c > (abc)^{\frac{1}{2}}$$

Also, because $\frac{1}{2}+\frac{1}{3}+\frac{1}{7} < 1$, we have:

$$1 > abc\Rightarrow 1 > abc^{1-\frac{1}{2}-\frac{1}{3}-\frac{1}{7}}$$

Multiplying these four inequalities, it follows that:

$$abc > (abc)^{\frac{1}{7}} \cdot (abc)^{\frac{1}{3}}\cdot (abc)^{\frac{1}{2}} \cdot abc^{1-\frac{1}{2}-\frac{1}{3}-\frac{1}{7}} = abc$$

a contradiction.

Answer 2

Let our minimum be equal to $k$.

Thus, $$a^{42}\geq k^6,$$ $$b^{42}\geq k^{14},$$ $$c^{42}\geq k^{21}$$ and $$1\geq k.$$ Thus, $$(abc)^{42}\geq k^{6+14+21+1}$$ or $$abc\geq k.$$

Answer 3

For a positive $a$ and $x>1$, $$ min(a^x,1) \leq min(a,1) \leq a $$ Moreover, $$ [min(a,b,1)]^2 = min(a^2,b^2,1) \leq min(a,1) \, min(b,1) \, . $$ This result is immediately extended to the case of three or more variables. Apply the two equations above to obtain $$ min(a^n,b^n,...,h^n,1) \leq a b ... h $$ for $n>0$ variables $a...h$. The first equation tells you that $$ min(a^{n_1},b^{n_2},...,h^{n_n},1) \leq a b ... h $$ when $n_1...n_n\geq n$.

Answer 4

One can also grind through to get an answer:

If $a,b,c \ge 1$ or any of $a,b,c$ is zero then clearly the statement is true.

So we can suppose all are strictly positive and $\min (a,b,c) <1$.

If $a^7$ is the $\min$ then $a^{7 \over 5} \le b$ and $a^{7 \over 2} \le c$ so $a^7 = a a^{7 \over 5} a^{7 \over 2} a^{11 \over 10} \le a a^{7 \over 5} a^{7 \over 2} \le abc$.

The same analysis mutatis mutandis applies to $b,c$.